A sequence is said to be $d$-complete if every large integer is the sum of distinct integers from the sequence, none of which divide any other. This particular case of $d$-completeness was conjectured by Erdős and Lewin [ErLe96], who (among other related results) prove this when $a=3$, $b=5$, and $c=7$.

Characterizes a "snug" finite set of natural numbers: all elements are within a multiplicative factor $(1 + ε)$ of the minimum. Specifically, for a finite set $A$ and $ε > 0$, all $a ∈ A$ satisfy $a < (1 + ε) · min(A)$.

Predicate for pairwise coprimality of three integers. Requires all three input values to be pairwise coprime to each other.

Let $a, b, c$ be three integers which are pairwise coprime. Is every large integer the sum of distinct integers of the form $a^k b^l c^m$ ($k, l, m ≥ 0$), none of which divide any other?

Equivalently: is the set ${a^k b^l c^m : k, l, m \geq 0}$ d-complete?

Note: For this not to reduce to the two-integer case, we need the integers to be greater than one and distinct.

The prize of $250 is offered by Erdős in [Er97] and [Er97e] for a 'proof or disproof'.

The main problem was resolved in the affirmative by GPT 5.6 (prompted by Snyder).

This was formalized in Lean by Alexeev.

Erdős and Lewin [ErLe96] proved this conjecture when $a = 3$, $b = 5$, and $c = 7$.

A simpler case: the set of numbers of the form $2^k 3^l$ ($k, l ≥ 0$) is d-complete.

This was initially conjectured by Erdős in 1992, who called it a "nice and difficult" problem, but it was quickly proven by Jansen and others using a simple inductive argument:

If $n = 2m$ is even, apply the inductive hypothesis to $m$ and double all summands.

If $n$ is odd, let $3^k$ be the largest power of $3$ with $3^k ≤ n$, and apply the inductive hypothesis to $n - 3^k$ (which is even).

In [Er92b] Erdős makes the stronger conjecture (for $a=2$, $b=3$, and $c=5$) that, for any $\epsilon>0$, all large integers $n$ can be written as the sum of distinct integers $b_1<\cdots <b_t$ of the form $2^k3^l5^m$ where $b_t<(1+\epsilon)b_1$.